What is the Relationship Between Velocity and Distance for a Sprinter?

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SUMMARY

The discussion centers on the relationship between velocity and distance for a sprinter, specifically analyzing a graph that depicts velocity in meters per second (m/s) against distance in meters (m) over a 100m sprint. The velocity increases linearly from 0 m/s to 12 m/s over the first 25 meters, then stabilizes between 12 m/s and 11.8 m/s for the remaining distance. Participants emphasize the need for calculus to derive average acceleration over the initial 15 meters and to approximate distance traveled within the first 5 seconds, given the non-constant acceleration during this period.

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  • Understanding of basic kinematics, including velocity and acceleration.
  • Familiarity with graphing concepts, particularly velocity vs. distance.
  • Knowledge of calculus, specifically integration techniques.
  • Ability to perform calculations related to motion, such as time taken and average acceleration.
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  • Learn calculus integration techniques to relate distance and time in motion equations.
  • Study kinematic equations for uniformly accelerated motion.
  • Explore the concept of average acceleration and how to calculate it over varying intervals.
  • Investigate graphing velocity vs. time and its implications in physics.
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Students and educators in physics, coaches analyzing sprinter performance, and anyone interested in the mathematical modeling of motion dynamics.

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y axis: velocity m/s
x axis: distance m

Graph models a sprinter over 100m

velocity is increasing in a straight line up to 25m from 0m/s to 12m/s, then is consistently between 12m/s and 11.8m/s for the remaining 75m.

I can calculate variables (time taken, average acceleration etc) for the remaining 75m or 50m as velocity isn't changing much.

I'm struggling to calculate the average acceleration over the first 15m and the approximate distance traveled during the first 5 seconds.

Thanks in advance for any help.
 
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Welcome to PF!
It is highly unusual to graph velocity vs distance; we nearly always graph velocity vs time.
Are you sure it is velocity vs distance?
If it is, you will need calculus to work it out as the acceleration will not be constant over the first 12 seconds.
The straight line on v vs x means v = kx or dx/dt = kx.
Change it to dx/x = k*dt
Integrate both sides to get a relation between x and t.
Then you can find velocity dx/dt and acceleration dv/dt.
 
hi thanks for the welcome, and thanks for the reply!

Yes it is definitely a velocity vs distance.

Thanks again for your reply.
 

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